Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 6.5 Curl and Divergence (13.5)
Def: Let F = P,Q, R be a vector field on  3 and partial derivatives of P,Q,R all exists, then -­‐ the curl of F is defined as
⎛ i
⎜
d
curlF = ∇ × F = det ⎜
⎜ dx
⎜
⎝ P
j
d
dy
Q
k ⎞
⎟
d ⎟
∂R ∂Q ∂P ∂R ∂Q ∂P
=
−
,
−
,
−
dz ⎟
∂y ∂z ∂z ∂x ∂x ∂y
⎟
R ⎠
-­‐ the divergence of F is defined as
∂ ∂ ∂
∂P ∂Q ∂R
, ,
⋅ P,Q, R =
+
+
∂x ∂y ∂z
∂x ∂y ∂z

Ex 8.
Let f ( x, y, z ) = x sin yz , then F = ∇f = sin yz, xz cos yz, xy cos yz and
∂P ∂Q ∂R
+
+
= 0 + −xz 2 sin yz + −xy 2 sin yz = −x sin yz y 2 + z 2 = − f ( x, y, z ) y 2 + z 2 divF =
∂x ∂y ∂z
∂R
∂Q ∂R
∂P ∂Q
∂P
= x cos yz − xyz sin yz =
,
= y cos yz =
,
= z cos yz =
and , so curlF = 0 . ∂y
∂z ∂x
∂z ∂x
∂y
Thm: Suppose f(x,y,z) has continuous second-­‐order partial derivatives, then curl ( ∇f ) = curl fx , fy , fz = fzy − fyz , fxz − fzx , fyx − fxy = 0 divF = ∇ ⋅ F =
(
) (
)
(
)
(
)
Thm: If F is vector field defined on  3 whose component functions have continuous partial derivatives and curlF = 0 , then F is a conservative vector field. Thm: Suppose F = P,Q, R is a vector field on  3 and has continuous second-­‐order partial derivatives, then ⎛ ∂R ∂Q ⎞ ⎛ ∂P ∂R ⎞ ⎛ ∂Q ∂P ⎞
∂R ∂Q ∂P ∂R ∂Q ∂P
divcurlF = div
−
,
−
,
−
=⎜
−
+
−
+
−
= 0 ⎝ ∂yx ∂zx ⎟⎠ ⎜⎝ ∂zy ∂xy ⎟⎠ ⎜⎝ ∂xz ∂yz ⎟⎠
∂y ∂z ∂z ∂x ∂x ∂y
Green’s Theorem (vector form): Let F = P,Q,0 = Pi + Qj + 0k ⎛ d d d
⎞
∂Q ∂P
, ,
× P,Q,0 ⎟ ⋅ 0,0,1 dA = ∫∫
−
dA = 
∫ F ⋅ dr ∂x ∂y
⎝ dx dy dz
⎠
D
C
∫∫ ( curlF ) ⋅ k dA = ∫∫ ⎜
D
D
x ' ( t ) , y' ( t )
y' ( t ) ,−x ' ( t )
T'

Thm: Let r ( t ) = x ( t ) , y ( t ) then T ( t ) =
and n ( t ) =
=N=


r '(t )
r '(t )
T'

Thm: Let F = P,Q be a vector field and r ' ( t ) = x ' ( t ) , y' ( t ) b
⎛
y' ( t ) ,−x ' ( t ) ⎞ 
F
⋅
n
ds
=
C∫
∫a ⎜⎝ P ( x (t ), y (t )),Q ( x (t ), y (t )) ⋅ r '(t ) ⎟⎠ r '(t ) dt =
⎛ ∂P ∂Q ⎞
∫a P ( x (t ), y (t )) y'(t ) − Q ( x (t ), y (t )) x '(t ) dt = ∫a P dy − Q dx = ∫∫D ⎜⎝ ∂x + ∂y ⎟⎠ dA = ∫∫D divF ( x, y ) dA
b
b
84 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 6.6 Surface Integrals (13.6)
The relationship of surface integral and surface area is analogical to the relationship between the line integral and arc length. 
Def: Let S be a surface r ( u,v ) = x ( u,v ) , y ( u,v ) , z ( u,v ) ,
(u,v ) ∈D . For a rectangular D the surface integral is defined by ∫∫
f ( x, y, z ) dS = lim
m,n→∞
S
∑ ∑ f ( P ) ΔS
m
n
*
ij
ij
i=1 j=1
where Pij* is a sample point on the patch Sij which area is ΔSij ≈ ru × rv ΔuΔv . For a non rectangular domain ∫∫ f ( x, y, z ) dS = ∫∫ f ( r ( u,v )) ru × rv dA S
D
Example: for z = g ( x, y ) we have ∫∫ f ( x, y, z ) dS = ∫∫ f ( x, y, g ( x, y ))
S
( g x )2 + ( g y )
2
+ 1 dA D
Def: A two-­‐sided surface S is called oriented if the unit normal vector n is defined at every point (except the boundary points). The orientation is chosen by direction (positive or negative) of the unit normal, in other words by choosing the side. Example: A Mobius strip is one-­‐sided surface, therefore non orientable. A torus is two-­‐sided, so it can be oriented inward or outward. Def: A closed surface is a boundary of solid region. A closed surface is considered positive oriented if the unit normal points outward. Def: Flux of F across surface S is defined by 
∫∫ F ⋅ dS = 
∫∫ F ⋅ n dS . Note the S
S
difference between dS and dS and the similarity with dr . Corollary:  
 

 
 
ru × rv
ru × rv
  dS =
 
F
⋅
dS
=
F
⋅
n
dS
=
F
⋅
F
r
u,v
⋅
ru × rv du dv = ∫∫ F ⋅ ( ru × rv ) du dv (
)
(
)
ru × rv

∫∫

∫∫

∫∫ ru ×rv
∫∫
S
S
S
D
(
)
D
Examples: 1) For z=g(x,y) one get 
∫∫ F ⋅ dS = ∫∫ P,Q, R ⋅ −gx ,−gy ,1 dA = ∫∫ −Pgx − Qgy + R dA S
D
D

2) Rate of flow of a fluid with density ρ and velocity field v is given by 
∫∫ ρv ⋅ n dS . S
3) Electric flux of an electric field E through the surface S is given by 
∫∫ E ⋅ dS . Furthermore, S
Gauss’s Law says that a net charge enclosed by a closed surface S is Q = ε 0 
∫∫ E ⋅ dS where ε 0 is S
a permittivity of free space. 4) Given a conductivity constant K of a substance and a temperature of a body given by u ( x, y, z ) , the rate of heat flow is given by −K∇u and the rate of heat flow by −K 
∫∫ ∇u ⋅ dS . S
85 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 
Ex 9. Let F = x, y, z and r (ϕ ,θ ) = sinϕ cosθ ,sinϕ sinθ ,cosϕ ,0 ≤ ϕ ≤ π ,0 ≤ θ ≤ 2π

r (ϕ ,θ ) = sin ϕ cosθ ,sin ϕ sin θ ,cosϕ

rϕ = cosϕ cosθ ,cosϕ sin θ ,− sin ϕ

rθ = − sin ϕ sin θ ,sin ϕ cosθ ,0

 
  
F ( r (ϕ ,θ )) ⋅ rϕ × rθ = r ⋅ rϕ × rθ =
(
)
(
)
sin ϕ cosθ ,sin ϕ sin θ ,cosϕ ⋅ ( cosϕ cosθ ,cosϕ sin θ ,− sin ϕ × − sin ϕ sin θ ,sin ϕ cosθ ,0 ) = sin ϕ
2π π

∫∫ F ⋅ dS = 
∫∫ sinϕ dA = ∫ ∫ sinϕ dϕ dθ = − 2π cosϕ
S
S
π
0
0 0
86 = −2π ( −1− 1) = 4π